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stellarik [79]
3 years ago
12

What is newton’s first law of motion ?

Physics
1 answer:
jolli1 [7]3 years ago
7 0

Newtons first law of motion states that an object will remain resting or in uniform motion in a straight line unless acted upon by an external force. It may be seen as a statement; that objects will remain in their state of motion unless a force acts to change the motion.

Hope this helps!

You might be interested in
What are 3 facts you learned about the periodic table:
d1i1m1o1n [39]

Answer: the rarest element is Francium. J is not on the periodic table. also Dmitri Mendeleev proposed the periodic table.

Explanation: Kinda looked the last one up.

7 0
3 years ago
A block of iron at 33.0°C has mass 2.30 kg. If 3.50×104 J of heat are transferred to the block, what is its resulting temperatur
spayn [35]

Answer:

66.8°C

Explanation:

dQ = m*cp*dT where:

m = mass of block

cp = specific heat of iron

dT = temperature change

-------

dQ = 3.5 10^4 J

cp = 0.45 Kj / Kg = 450 J / Kg

3.5 10^4 = 2.3 * 450 * dt-----dt = 35000 / 450 * 2.3 = 33.8 °

-------

Final temperature = 33 + 33.8 = 66.8 °C

3 0
3 years ago
A radio signal has a frequency of 1.023 x 108 HZ. If the speed of the signal in air is 2.997 x 108m/s, what is the wavelength of
Sladkaya [172]

Answer:

2.93 m  (which agrees with answer "C" on the list)

Explanation:

Recall that the speed of the wave equals the product of the wave's length times its frequency. Therefore, the wavelength is going to be the quotient of the speed of the signal divided its frequency:

Wavelength = 2.997  10^8 / 1.023  10^8 =  2.93 m

5 0
3 years ago
Three moles of an ideal gas are taken around the eyele acb shown in Fig.For this gas Co-29.2J/mol-K.Process ac is at constant pr
Scorpion4ik [409]

For three moles of an ideal gas are taken around the eye, the total work for the cycle is mathematically given as

Wt=1945.475J

<h3>What is the total work F for the cycle.(R=8.31J/molK)?</h3>

Generally, the equation for work  is mathematically given as

Wt=wac+Wc+Wba

Therefore

Wac=Pa(Vc-Va)=nR(Tc-Ta)

Wac=3(8.314*192)

Wac=4788.864J

Wcb=P1v1-p2v2/v-1

Wcb-3*8.34*108/0.4

Wcb=-6734.34J

Wab=0

In conclusion,

Wt=wac+Wc+Wba

Wt=4788.864J+0+(-6734.34J)

Wt=+1945.475J

Read more about work

brainly.com/question/756198

#SPJ1

8 0
2 years ago
What is the terminal velocity of a 6.00-kg mass object in falling with a drag force with a magnitude that depends on speed, v, a
3241004551 [841]

When object reached the terminal speed then its acceleration is zero

So as per Newton's II law we can say

F_{net} = 0

now in that case we can say that net force is zero so here weight of the object is counter balanced by the drag force when it will reach at terminal speed

so we can write

mg - F_d = 0

so here we are given that

F_d = 30

6*9.8 - 30*v = 0

58.8 - 30 *v = 0

v = \frac{58.8}{30}

v = 1.96 m/s

so terminal speed will be nearly 2 m/s

6 0
3 years ago
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